Question

Match the two columns

Column 1	Column 2
$6.\overline{54}$ is ……..	$14$
$\mathrm{\pi }$ is	$6$
The length of period of $\frac{1}{7}=$….	a rational number
If $x=2-\sqrt{3}$, then $\left(x^2+\frac{1}{x^2}\right)=$……	an irrational number

Answer 1

For (A)-(iii)

Let's simplify $6.\overline{54}$

Let $x=6.5454....$ and $100x=654.5454....$

$$\begin{gathered} 100x-x=654.5454...-6.5454... \\ 99x=648 \\ x=\frac{648}{99} \end{gathered}$$

Hence, $x$ is a rational number as it is of the form $\frac{p}{q},p,q$are integers and $q\ne 0$.

Hence, $x$ is a rational number, is the correct answer.

For (B)-(iv)

The value of $\mathrm{\pi }$ is $3.141592...$

Since the value of $\mathrm{\pi }$ is non-terminating and non-recurring, it is an irrational number.

Hence, $\mathrm{\pi }$ is an irrational number is the correct answer.

For (C)-(ii)

The value of $\frac{1}{7}=0.\overline{142857}$

The repeating part to the right of the decimal point has six numbers, which is the period.

Hence, the length of the period of $\frac{1}{7}$ is $6$ is the correct answer.

For (D)-(ii)

$x=2-\sqrt{3}$ then

$\frac{1}{x}=\frac{1}{2-\sqrt{3}}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$ [$\frac{1}{x}$is rationalized]

$$\begin{gathered} \left(x^2+\frac{1}{x^2}\right)={\left(2-\sqrt{3}\right)}^2+{\left(2+\sqrt{3}\right)}^2 \\ =2^2-4\sqrt{3}+{\left(\sqrt{3}\right)}^2+2^2+4\sqrt{3}+{\left(\sqrt{3}\right)}^2 \\ =4+4+3+3=14 \end{gathered}$$ [$$\begin{gathered} {\left(a+b\right)}^2=a^2+2ab+b^2 \\ {\left(a-b\right)}^2=a^2-2ab+b^2 \end{gathered}$$is used]

Hence, $14$ is the correct answer.

Hence, the final answer is (A)-(iii),(B)-(iv),(C)-(ii),(D)-(ii)